Pseudo-Isotropic Centro-Affine Lorentzian Surfaces

Abstract

In this paper, we study centro-affine Lorentzian surfaces $M^2$ in ${ℝ}^3$ which have pseudo-isotropic or lightlike pseudo-isotropic difference tensor. We first show that $M^2$ is pseudo-isotropic if and only if the Tchebychev form $T=0$. In that case, $M^2$ is a an equi-affine sphere. Next, we will get a complete classification of centro-affine Lorentzian surfaces which are lightlike pseudo-isotropic but not pseudo-isotropic.

1. Introduction

The notion of isotropic submanifolds of a Riemannian manifold was first introduced by B. O’Neill [1] who studied submanifolds for which the second fundamental form is isotropic. This notion has recently be extended by Cabrerizo, Fernández and Gómez in [2] for pseudo-Riemannian manifolds. Such manifolds are called pseudo-isotropic. Pseudo-isotropic Lagrangian surfaces has been studied in [3]. This notion of isotropy can be extended to all bundle valued tensor fields $\tilde{T}$ by saying that $\tilde{T}$ is isotropic if and only if the value of $g(\tilde{T}(V,\cdots ,V),\tilde{T}(V,\cdots ,V))$ is independent of the unit vector V. In the affine case, hypersurfaces with isotropic difference tensor K have been studied in [4,5,6].

In this paper, we study centro-affine Lorentzian surfaces $M^2$ in ${ℝ}^3$. The basic notions of centro-affine geometry are recalled in Section 2. A remarkable fact about affine hypersurfaces is that we can both introduce an affine induced connection ∇ and a Levi–Civita connection $\widehat{\nabla }$ of the affine metric h. In general those two connections do not coincide. The geometry of the hypersurface is determined by the difference tensor K defined by $K(X,Y)={\nabla }_{X}Y-{\widehat{\nabla }}_{X}Y$.

We say that $M^2$ is pseudo-istropic if for any point p and tangent vector V at p we have

$$h(K(V,V),K(V,V))=\lambda \left(p\right)h^2(V,V),$$

(1)

where $\lambda$ is a function on $M^2$.

Since $M^2$ is equipped with a Lorentzian metric, we can consider lightlike, timelike and spacelike vectors. We say that $M^2$ is timelike (resp. spacelike) pseudo-isotropic if the Equation (1) is satisfied for any point p and any timelike (resp. spacelike) vector V at p. We say that $M^2$ is lightlike pseudo-isotropic if for any point p and lightlike vector V at p, $K(V,V)$ is a lightlike vector.

Note that some results in ([2]) remain valid if we replace the second fundamental form by K. So the following conditions are equivalent

• $M^2$ is pseudo-isotropic;
• $M^2$ is timelike pseudo-isotropic;
• $M^2$ is spacelike pseudo-isotropic.

We first show that

Theorem 1.

Let $M^2$ be a centro-affine Lorentzian surface in ${ℝ}^3$. Then $M^2$ is pseudo-isotropic if and only if $T\left(X\right)={\displaystyle \frac{1}{2}}Tr{K}_X=0$ for all tangent vector field X.

Then, in Section 4, we prove the following theorem

Theorem 2.

Let $M^2$ be a centro-affine Lorentzian surface in ${ℝ}^3$, which is lightlike pseudo-isotropic but not pseudo-isotropic. Then $M^2$ is locally congruent with one of the following immersions

1. 

$\mathcal{F}=(C_{1}exp\left(v\right)+C_{2}exp(-v)+C_3)exp\left(u\right)+C_{1}exp\left(v\right)$;

2. 

$\mathcal{F}=(C_{1}exp(-v)+C_2)exp\left(u\right)+C_{3}exp\left(v\right)-C_{1}exp(-v)-2C_2$;

3. 

$\mathcal{F}=\left(C_1\left({\displaystyle \frac{1+\sqrt{1-r^2}}{r}}\right)exp\left(\sqrt{1-r^2}v\right)+C_2\left({\displaystyle \frac{1-\sqrt{1-r^2}}{r}}\right)exp(-\sqrt{1-r^2}v)+{\displaystyle \frac{C_3}{r}}\right)exp\left(u\right)+C_{1}exp\left(\sqrt{1-r^2}v\right)+C_{2}exp(-\sqrt{1-r^2}v)+C_3$;

4. 

$\mathcal{F}=\left({\displaystyle \frac{C_1-C_2\sqrt{r^2-1}}{r}}\right(cos\left(\sqrt{r^2-1}v\right)+\left({\displaystyle \frac{C_2+C_1\sqrt{r^2-1}}{r}}\right)sin\left(\sqrt{r^2-1}v\right)+{\displaystyle \frac{C_3}{r}})exp\left(u\right)+C_{1}cos\left(\sqrt{r^2-1}v\right)+C_{2}sin\left(\sqrt{r^2-1}v\right)+C_3$ ;

5. 

$\mathcal{F}=(C_1{\displaystyle \frac{v^2}{2}}+(C_2-C_1)v+(C_3-C_2))exp\left(u\right)+C_1{\displaystyle \frac{v^2}{2}}+C_{2}v+C_3$;

6. 

$\mathcal{F}=\left(C_1\right({\displaystyle \frac{\sqrt{1+r^2}-1}{r}})exp\left(\sqrt{1+r^2}v\right)-C_2({\displaystyle \frac{\sqrt{1+r^2}+1}{r}})exp(-\sqrt{1+r^2}v)-{\displaystyle \frac{C_3}{r}})exp\left(u\right)+C_{1}exp\left(\sqrt{1+r^2}v\right)+C_{2}exp(-\sqrt{1+r^2}v)+C_3$.

where $C_1=(1,0,0)$, $C_2=(0,1,0)$ and $C_3=(0,0,1)$ and r is a strictly positive constant.

2. Preliminary

Here we first recall the basic notions of affine differentiable geometry and centro-affine hypersurfaces. For more details we refer to [7,8]. Let $f:M^n\to {ℝ}^{n+1}$ be an immersion of a connected differentiable n-dimensional manifold into the affine space ${ℝ}^{n+1}$ equipped with its usual flat connection D and let $\xi$ be an arbitrary local transversal vector field to $f\left(M^n\right)$.

For any vector fields $X,Y$ we have

$$\begin{array}{ccc}\hfill D_X{f}_{*}\left(Y\right)& =& f_{*}\left({\nabla }_{X}Y\right)+h(X,Y)\xi ,\hfill \end{array}$$

(2)

$$\begin{array}{ccc}\hfill D_X\xi & =& -f_{*}\left(SX\right).\hfill \end{array}$$

(3)

where ∇ is a torsion-free connection of $M^n$ called the affine induced connection, h a non degenerate $(0,2)$-tensor field called the affine metric and S a (1-1)-tensor field called the affine shape operator.

The hypersurface $M^n$ is said to be a centro-affine hypersurface if the position vector is nowhere tangent to $M^n$. Then f is a transversal field along itself. We can take $\xi =-f$ and the Equation (3) becomes $D_X\xi =-f_{*}\left(X\right)$.

The following fundamental equations of Gauss and Codazzi are given by

$$\begin{array}{c}R(X,Y)Z=h(Y,Z)X-h(X,Z)Y\left(\mathrm{Equation}\mathrm{of}\mathrm{Gauss}\right),\hfill \end{array}$$

(4)

$$\begin{array}{c}(\nabla h)(X,Y,Z)=(\nabla h)(Y,X,Z)\left(\mathrm{Equation}\mathrm{of}\mathrm{Codazzi}\mathrm{for}h\right).\hfill \end{array}$$

(5)

where X, Y and Z are vector fields on $M^n$.

We denote by $\widehat{\nabla }$ the Levi–Civita connection of the centro-affine metric h and by $\widehat{R}$ the curvature tensor of $\widehat{\nabla }$. The difference tensor K is defined by

$$K(X,Y)={\nabla }_{X}Y-{\widehat{\nabla }}_{X}Y,$$

for vector fields X and Y on $M^n$. We also write $K_{X}Y=K(X,Y)$ and $K_X={\nabla }_X-{\widehat{\nabla }}_X$. Thus, for each X, it follows that $K_X$ is a tensor of type $(1,1)$ that maps Y to $K(X,Y)$. Since both ∇ and $\widehat{\nabla }$ have zero torsion, K is symmetric in X and Y

$$K(X,Y)=K(Y,X).$$

(6)

From (5) it follows that $h\left(K\right(X,Y),Z)$ is symmetric in X, Y and Z.

On the other hand expressing the Gauss equation in terms of the Levi–Civita connection of the centro-affine metric, we deduce that

$$\widehat{R}(X,Y)Z=K_Y{K}_{X}Z-K_X{K}_{Y}Z+(h(Y,Z)X-h(X,Z)Y).$$

(7)

We also obtain a Codazzi equation for K

$$\left({\widehat{\nabla }}_{X}K\right)(Y,Z)=\left({\widehat{\nabla }}_{Y}K\right)(X,Z).$$

(8)

The Tchebychev form T is defined by

$$T\left(X\right)={\displaystyle \frac{1}{n}}trace{K}_X.$$

3. Equiaffine Sphere and Pseudo-Isotropy

Now we suppose that $M^2$ is a Lorentzian centro-affine pseudo-isotropic surface in ${ℝ}^3$. The surface is equipped with an affine indefinite metric h. Since $M^2$ is pseudo-isotropic, as in the paper ([5]), by linearization of (1), we obtain at any point p on $M^2$

$$\begin{array}{ll}h\left(K\right(X,Y),K(Z,V\left)\right)& +h\left(K\right(X,Z),K(V,Y\left)\right)+h\left(K\right(X,V),K(Y,Z\left)\right)\\ & =\lambda \left(p\right)\left(h\right(X,Y\left)h\right(Z,V)+h(X,Z\left)h\right(V,Y)+h(X,V\left)h\right(Z,Y\left)\right).\end{array}$$

(9)

Let $p\in M^2$, we say that $\{E_1,E_2\}$ is a null frame at p if it satisfies $h(E_i,E_j)=1-{\delta }_{ij}$, $i,j\in \{1,2\}$. From (1), we find immediately that $h(K(E_1,E_1),K(E_1,E_1))=h(K(E_2,E_2),K(E_2,E_2))=0$ so $K(E_1,E_1)$ and $K(E_2,E_2)$ are lightlike vectors.

Using (9) with $X=Y=Z=E_1$ and $V=E_2$ (resp. $X=Y=Z=E_2$ and $V=E_1$), we get that $K(E_1,E_2)$ is orthogonal to $K(E_1,E_1)$ (resp. $K(E_2,E_2)$).

We can write $K(E_1,E_2)={\alpha }_1{E}_1+{\alpha }_2{E}_2$. Using the symmetry of $h\left(K\right(X,Y),Z)$ in X, Y and Z and the fact that $\{E_1,E_2\}$ is a null frame, we prove that

$$\begin{array}{c}K(E_1,E_1)={\alpha }_2{E}_1+{\alpha }_3{E}_2,\hfill \\ K(E_2,E_2)={\alpha }_4{E}_1+{\alpha }_1{E}_2.\hfill \end{array}$$

Since $K(E_1,E_1)$ and $K(E_2,E_2)$ are lightlike vectors, we obtain ${\alpha }_2{\alpha }_3=0$ and ${\alpha }_1{\alpha }_4=0$. Using the fact that $K(E_1,E_2)$ is orthogonal to $K(E_1,E_1)$ and $K(E_2,E_2)$, we get

$$\begin{array}{c}{\alpha }_1{\alpha }_3+{\alpha }_2^2=0,\hfill \\ {\alpha }_2{\alpha }_4+{\alpha }_1^2=0.\hfill \end{array}$$

Therefore, we have ${\alpha }_1={\alpha }_2=0$ so $K_{E_1}=\left(\begin{array}{cc}0& 0\\ {\alpha }_3& 0\end{array}\right)$ and $K_{E_2}=\left(\begin{array}{cc}0& {\alpha }_4\\ 0& 0\end{array}\right)$. We have proved that $Tr{K}_{E_1}=Tr{K}_{E_2}=0$. For any vector X at a point p, we show that $h(K(X,E_1),E_1)=h(K(X,E_2),E_2)=0$ so $Tr{K}_X=0$.

Conversely, suppose that $M^2$ is a Lorentzian centro-affine surface in ${ℝ}^3$ with $Tr{K}_X=0$. As before, we have

$$\begin{array}{c}K(E_1,E_2)={\alpha }_1{E}_1+{\alpha }_2{E}_2,\hfill \\ K(E_1,E_1)={\alpha }_2{E}_1+{\alpha }_3{E}_2,\hfill \\ K(E_2,E_2)={\alpha }_4{E}_1+{\alpha }_1{E}_2.\hfill \end{array}$$

From $Tr{K}_{E_1}=Tr{K}_{E_2}=0$, we deduce that ${\alpha }_1={\alpha }_2=0$.

Let X a vector at a point p, we can write $X={\beta }_1{E}_1+{\beta }_2{E}_2$ and computations give

$$\begin{array}{cc}\hfill h\left(K\right(X,X),K(X,X\left)\right)& =2{\beta }_1^2{\beta }_2^2{\alpha }_3{\alpha }_4,\hfill \\ \hfill h{(X,X)}^2& =4{\beta }_1^2{\beta }_2^2.\hfill \end{array}$$

So $h(K(X,X),K(X,X))={\displaystyle \frac{{\alpha }_3{\alpha }_4}{2}}h{(X,X)}^2$. Since $h\left(K\right(X,X),K(X,X\left)\right)$ and $h{(X,X)}^2$ are independent of the basis, it is the same for ${\displaystyle \frac{{\alpha }_3{\alpha }_4}{2}}$. Therefore $M^2$ is pseudo-isotropic. This completes the proof of Theorem 1.

4. Lightlike Pseudo-Isotropic Lorentzian Surfaces

Now we suppose that $M^2$ is proper lightlike pseudo-isotropic but not pseudo-isotropic. We say that $M^2$ is lightlike pseudo-isotropic of Type 1 at the point p if there exists a lightlike vector V such that $K(V,V)$ and V are independent and of Type 2 if for every lightlike vector V, $K(V,V)$ and V are dependent. The points of Type 1 form an open set $U_1$ of $M^2$ and the points of Type 2 a closed set $U_2$. It is clear that the null frame $\{E_1,E_2\}$ can be extended differentially to a neighborhood of a point p belonging either to $U_1$ or the interior of $U_2$. Given the structure of surfaces of this section, it follows that the classification theorem remain valid on the whole of $M^2$.

Lemma 1.

There are no lightlike pseudo-isotropic surfaces of Type 1.

Proof. 

We consider a null frame in a neighborhood of p constructed as before. We take $E_1$ the lightlike vector such that $K(E_1,E_1)$ and $E_1$ are independent. Then $K(E_1,E_1)$ is a multiple of $E_2$ and we can choose the frame such that $K(E_1,E_1)=E_2$. We write $K(E_2,E_2)=\beta E_1+\alpha E_2$. Using the symmetry of $h\left(K\right(X,Y),Z)$ and the fact that $\{E_1,E_2\}$ is a null frame, we get $K(E_1,E_2)=\alpha E_1$. As $E_2$ is lightlike, we must have that $K(E_2,E_2)$ is also lightlike and we obtain $\alpha \beta =0$. From $Tr{K}_{E_1}\ne 0$, we get $\alpha \ne 0$ so $\beta =0$. Therefore, we have

$$\begin{array}{c}K(E_1,E_1)=E_2,\hfill \\ K(E_1,E_2)=\alpha E_1,\hfill \\ K(E_2,E_2)=\alpha E_2.\hfill \end{array}$$

As $\widehat{\nabla }$ is the Levi–Civita connection, we have $X.h(Y,Z)=h({\widehat{\nabla }}_{X}Y,Z)+h(Y,{\widehat{\nabla }}_{X}Z)$ for all vector fields on $M^2$. Using the previous expression with $X=Y=Z=E_1$ and $X=Y=Z=E_2$, we prove that ${\widehat{\nabla }}_{E_1}{E}_1=\gamma E_1$ and ${\widehat{\nabla }}_{E_2}{E}_2=\delta E_2$ where $\gamma$ and $\delta$ are functions. Using the same expression, by replacing $X,Y,Z$ with $E_1$ or $E_2$, we get ${\widehat{\nabla }}_{E_1}{E}_2=-\gamma E_2$ and ${\widehat{\nabla }}_{E_2}{E}_1=-\delta E_1$. We have

$$\begin{array}{cc}\hfill \widehat{\nabla }K(E_2,E_1,E_1)& ={\widehat{\nabla }}_{E_2}K(E_1,E_1)-2K({\widehat{\nabla }}_{E_2}{E}_1,E_1)\hfill \\ \hfill & ={\widehat{\nabla }}_{E_2}{E}_2-2K(-\delta E_1,E_1)\hfill \\ \hfill & =\delta E_2+2\delta E_2\hfill \\ \hfill & =3\delta E_2,\hfill \end{array}$$

$$\begin{array}{cc}\hfill \widehat{\nabla }K(E_1,E_2,E_1)& ={\widehat{\nabla }}_{E_1}K(E_2,E_1)-K({\widehat{\nabla }}_{E_1}{E}_2,E_1)-K(E_2,{\widehat{\nabla }}_{E_1}{E}_1)\hfill \\ \hfill & ={\widehat{\nabla }}_{E_1}\left(\alpha E_1\right)-K(-\gamma E_2,E_1)-K(\gamma E_1,E_2)\hfill \\ \hfill & =(E_1\left(\alpha \right)+\alpha \gamma )E_1,\hfill \end{array}$$

$$\begin{array}{cc}\hfill \widehat{\nabla }K(E_1,E_2,E_2)& ={\widehat{\nabla }}_{E_1}K(E_2,E_2)-2K({\widehat{\nabla }}_{E_1}{E}_2,E_2)\hfill \\ \hfill & ={\widehat{\nabla }}_{E_1}\left(\alpha E_2\right)-2K(-\gamma E_2,E_2)\hfill \\ \hfill & =(E_1\left(\alpha \right)+\alpha \gamma )E_2,\hfill \end{array}$$

$$\begin{array}{cc}\hfill \widehat{\nabla }K(E_2,E_1,E_2)& ={\widehat{\nabla }}_{E_2}K(E_1,E_2)-K({\widehat{\nabla }}_{E_2}{E}_1,E_2)-K(E_1,{\widehat{\nabla }}_{E_2}{E}_2)\hfill \\ \hfill & ={\widehat{\nabla }}_{E_2}\left(\alpha E_1\right)-K(-\delta E_1,E_2)-K(E_1,\delta E_2)\hfill \\ \hfill & =(E_2\left(\alpha \right)-\alpha \delta )E_1.\hfill \end{array}$$

So computations give

$$\begin{array}{c}\widehat{\nabla }K(E_2,E_1,E_1)=3\delta E_2,\hfill \\ \widehat{\nabla }K(E_1,E_2,E_1)=(E_1\left(\alpha \right)+\alpha \gamma )E_1,\hfill \\ \widehat{\nabla }K(E_1,E_2,E_2)=(E_1\left(\alpha \right)+\alpha \gamma )E_2,\hfill \\ \widehat{\nabla }K(E_2,E_1,E_2)=(E_2\left(\alpha \right)-\alpha \delta )E_1.\hfill \end{array}$$

Then, from the Codazzi equation (8), we find that $\delta =0$ and we see that $\alpha$ satisfies the following system of differential equations $E_1\left(\alpha \right)=-\alpha \gamma$ and $E_2\left(\alpha \right)=0$.

If we compute $[E_1,E_2]\left(\alpha \right)$ in two different ways, we have

$$\begin{array}{cc}\hfill [E_1,E_2]\left(\alpha \right)& =E_1\left(E_2\left(\alpha \right)\right)-E_2\left(E_1\left(\alpha \right)\right)\hfill \\ \hfill & =0-E_2(-\alpha \gamma )\hfill \\ \hfill & =\alpha E_2\left(\gamma \right),\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill [E_1,E_2]\left(\alpha \right)& =({\widehat{\nabla }}_{E_1}{E}_2-{\widehat{\nabla }}_{E_2}{E}_1)\left(\alpha \right)\hfill \\ \hfill & =-\gamma E_2\left(\alpha \right)\hfill \\ \hfill & =0.\hfill \end{array}$$

Therefore $E_2\left(\gamma \right)=0$.

It is well known that $\widehat{R}(X,Y)Z={\widehat{\nabla }}_X{\widehat{\nabla }}_{Y}Z-{\widehat{\nabla }}_Y{\widehat{\nabla }}_{X}Z-{\widehat{\nabla }}_{[X,Y]}Z$, so

$$\begin{array}{cc}\hfill \widehat{R}(E_1,E_2)E_1& =0-{\widehat{\nabla }}_{E_2}{\widehat{\nabla }}_{E_1}{E}_1-{\widehat{\nabla }}_{({\widehat{\nabla }}_{E_1}{E}_2-0)}{E}_1\hfill \\ \hfill & =-{\widehat{\nabla }}_{E_2}\left(\gamma E_1\right)-{\widehat{\nabla }}_{(-\gamma E_2)}{E}_1\hfill \\ \hfill & =-E_2\left(\gamma \right)E_1\hfill \\ \hfill & =0.\hfill \end{array}$$

From (7), we get

$$\begin{array}{cc}\hfill \widehat{R}(E_1,E_2)E_1& =K_{E_2}{K}_{E_1}{E}_1-K_{E_1}{K}_{E_2}{E}_1+(h(E_2,E_1)E_1-0)\hfill \\ \hfill & =K_{E_2}{E}_2-K_{E_1}\left(\alpha E_1\right)+E_1\hfill \\ \hfill & =\alpha E_2-\alpha E_2+E_1\hfill \\ \hfill & =E_1.\hfill \end{array}$$

Hence, we obtain a contradiction. □

Now we deal with lightlike pseudo-isotropic surface of Type 2. We can choose a null frame such that $K(E_1,E_1)=E_1$ and $K(E_2,E_2)=\alpha E_2$. The symmetry of $h\left(K\right(X,Y),Z)$ gives $K(E_1,E_2)=\alpha E_1+E_2$. As before, we have ${\widehat{\nabla }}_{E_1}{E}_1=\gamma E_1$, ${\widehat{\nabla }}_{E_2}{E}_2=\delta E_2$, ${\widehat{\nabla }}_{E_1}{E}_2=-\gamma E_2$ and ${\widehat{\nabla }}_{E_2}{E}_1=-\delta E_1$ where $\gamma$ and $\delta$ are functions.

Lemma 2.

We have $\gamma =0$ and the functions α and δ satisfy the following system of differential equations

$$\begin{array}{c}{E}_1\left(\alpha \right)=\delta ,\hfill \\ E_2\left(\alpha \right)=\delta \alpha ,\hfill \\ E_1\left(\delta \right)=\alpha -1,\hfill \\ E_2\left(\delta \right)=\alpha (\alpha -1).\hfill \end{array}$$

Proof. 

We find

$$\begin{array}{cc}\hfill \widehat{\nabla }K(E_2,E_1,E_1)& ={\widehat{\nabla }}_{E_2}K(E_1,E_1)-2K({\widehat{\nabla }}_{E_2}{E}_1,E_1)\hfill \\ \hfill & ={\widehat{\nabla }}_{E_2}{E}_1-2K(-\delta E_1,E_1)\hfill \\ \hfill & =-\delta E_1+2\delta E_1\hfill \\ \hfill & =\delta E_1,\hfill \end{array}$$

$$\begin{array}{cc}\hfill \widehat{\nabla }K(E_1,E_2,E_1)& ={\widehat{\nabla }}_{E_1}K(E_2,E_1)-K({\widehat{\nabla }}_{E_1}{E}_2,E_1)-K(E_2,{\widehat{\nabla }}_{E_1}{E}_1)\hfill \\ \hfill & ={\widehat{\nabla }}_{E_1}(\alpha E_1+E_2)-K(-\gamma E_2,E_1)-K(\gamma E_1,E_2)\hfill \\ \hfill & =(E_1\left(\alpha \right)+\alpha \gamma )E_1-\gamma E_2,\hfill \end{array}$$

$$\begin{array}{cc}\hfill \widehat{\nabla }K(E_1,E_2,E_2)& ={\widehat{\nabla }}_{E_1}K(E_2,E_2)-2K({\widehat{\nabla }}_{E_1}{E}_2,E_2)\hfill \\ \hfill & ={\widehat{\nabla }}_{E_1}\left(\alpha E_2\right)-2K(-\gamma E_2,E_2)\hfill \\ \hfill & =(E_1\left(\alpha \right)+\alpha \gamma )E_2,\hfill \end{array}$$

$$\begin{array}{cc}\hfill \widehat{\nabla }K(E_2,E_1,E_2)& ={\widehat{\nabla }}_{E_2}K(E_1,E_2)-K({\widehat{\nabla }}_{E_2}{E}_1,E_2)-K(E_1,{\widehat{\nabla }}_{E_2}{E}_2)\hfill \\ \hfill & ={\widehat{\nabla }}_{E_2}(\alpha E_1+E_2)-K(-\delta E_1,E_2)-K(E_1,\delta E_2)\hfill \\ \hfill & =(E_2\left(\alpha \right)-\alpha \delta )E_1+\delta E_2.\hfill \end{array}$$

So computations give

$$\begin{array}{c}\widehat{\nabla }K(E_2,E_1,E_1)=\delta E_1,\hfill \\ \widehat{\nabla }K(E_1,E_2,E_1)=(E_1\left(\alpha \right)+\alpha \gamma )E_1-\gamma E_2,\hfill \\ \widehat{\nabla }K(E_1,E_2,E_2)=(E_1\left(\alpha \right)+\alpha \gamma )E_2,\hfill \\ \widehat{\nabla }K(E_2,E_1,E_2)=(E_2\left(\alpha \right)-\alpha \delta )E_1+\delta E_2.\hfill \end{array}$$

Then, from the Codazzi equation (8), we get $\gamma =0$ and we see that $\alpha$ satisfies the following system of differential equations $E_1\left(\alpha \right)=\delta$ and $E_2\left(\alpha \right)=\delta \alpha$.

If we compute $[E_1,E_2]\left(\alpha \right)$ in two different ways, we have

$$\begin{array}{cc}\hfill [E_1,E_2]\left(\alpha \right)& =E_1\left(\delta \alpha \right)-E_2\left(\delta \right)\hfill \\ \hfill & =E_1\left(\delta \right)\alpha +{\delta }^2-E_2\left(\delta \right),\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill [E_1,E_2]\left(\alpha \right)& =-{\widehat{\nabla }}_{E_2}{E}_1\left(\alpha \right)\hfill \\ \hfill & =\delta E_1\left(\alpha \right)\hfill \\ \hfill & ={\delta }^2.\hfill \end{array}$$

Therefore $E_2\left(\delta \right)=\alpha E_1\left(\delta \right)$.

From $\widehat{R}(X,Y)Z={\widehat{\nabla }}_X{\widehat{\nabla }}_{Y}Z-{\widehat{\nabla }}_Y{\widehat{\nabla }}_{X}Z-{\widehat{\nabla }}_{[X,Y]}Z$, we get

$$\begin{array}{cc}\hfill \widehat{R}(E_1,E_2)E_1& ={\widehat{\nabla }}_{E_1}{\widehat{\nabla }}_{E_2}{E}_1-0-{\widehat{\nabla }}_{(0-{\widehat{\nabla }}_{E_2}{E}_1)}{E}_1\hfill \\ \hfill & ={\widehat{\nabla }}_{E_1}(-\delta E_1)+{\widehat{\nabla }}_{(-\delta E_1)}{E}_1\hfill \\ \hfill & =-E_1\left(\delta \right)E_1.\hfill \end{array}$$

From (7), we get

$$\begin{array}{cc}\hfill \widehat{R}(E_1,E_2)E_1& =K_{E_2}{K}_{E_1}{E}_1-K_{E_1}{K}_{E_2}{E}_1+(h(E_2,E_1)E_1-0)\hfill \\ \hfill & =K_{E_2}\left(E_1\right)-K_{E_1}(\alpha E_1+E_2)+E_1\hfill \\ \hfill & =\alpha E_1+E_2-\alpha E_1-(\alpha E_1+E_2)+E_1\hfill \\ \hfill & =(1-\alpha )E_1.\hfill \end{array}$$

Therefore $E_1\left(\delta \right)=\alpha -1$ and $E_2\left(\delta \right)=\alpha (\alpha -1)$. □

It is easy to check that $E_1({(\alpha -1)}^2-{\delta }^2)=E_2({(\alpha -1)}^2-{\delta }^2)=0$. Therefore, we have the

Corollary 1.

There exist a constant $c\in \{-1,0,1\}$ and a non negative constant r such that

$c{r}^2={(\alpha -1)}^2-{\delta }^2$.

Lemma 3.

There exist local coordinates u and v such that

$$\begin{array}{cc}\hfill & \frac{\partial }{\partial u}=E_1,\hfill \\ \hfill & \frac{\partial }{\partial v}=E_2-\alpha E_1.\hfill \end{array}$$

Proof. 

If we define vector fields U and V by $U=E_1$ and $V=E_2-\alpha E_1$, we obtain

$$\begin{array}{cc}\hfill [U,V]& =[E_1,E_2-\alpha E_1]\hfill \\ \hfill & ={\widehat{\nabla }}_{E_1}{E}_2-{\widehat{\nabla }}_{E_2}{E}_1+{\widehat{\nabla }}_{E_1}(-\alpha E_1)-{\widehat{\nabla }}_{(-\alpha E_1)}{E}_1\hfill \\ \hfill & =0+\delta E_1-E_1\left(\alpha \right)E_1-0\hfill \\ \hfill & =0.\hfill \end{array}$$

□

It is easy to check that the previous systems of differential equations in terms of the coordinates $u,v$ reduce to

$$\begin{array}{cc}\hfill \frac{\partial }{\partial u}\left(\alpha \right)& =\delta ,\hfill \\ \hfill \frac{\partial }{\partial v}\left(\alpha \right)& =0,\hfill \\ \hfill \frac{\partial }{\partial u}\left(\delta \right)& =\alpha -1,\hfill \\ \hfill \frac{\partial }{\partial v}\left(\delta \right)& =0.\hfill \end{array}$$

If we denote the immersion of $M^2$ in ${ℝ}^3$ by $\mathcal{F}$, using (2), we get

$$\begin{array}{cc}\hfill {\mathcal{F}}_{uu}& =D_{\frac{\partial }{\partial u}}d\mathcal{F}\left(\frac{\partial }{\partial u}\right)=d\mathcal{F}\left({\tilde{\nabla }}_{\frac{\partial }{\partial u}}\frac{\partial }{\partial u}\right)+d\mathcal{F}\left(K(u,u)\right)-h(u,u)\mathcal{F},\hfill \\ \hfill {\mathcal{F}}_{uv}& =D_{\frac{\partial }{\partial u}}d\mathcal{F}\left(\frac{\partial }{\partial v}\right)=d\mathcal{F}\left({\tilde{\nabla }}_{\frac{\partial }{\partial u}}\frac{\partial }{\partial v}\right)+d\mathcal{F}\left(K(u,v)\right)-h(u,v)\mathcal{F},\hfill \\ \hfill {\mathcal{F}}_{vv}& =D_{\frac{\partial }{\partial v}}d\mathcal{F}\left(\frac{\partial }{\partial v}\right)=d\mathcal{F}\left({\tilde{\nabla }}_{\frac{\partial }{\partial v}}\frac{\partial }{\partial v}\right)+d\mathcal{F}\left(K(v,v)\right)-h(v,v)\mathcal{F}.\hfill \end{array}$$

We have the

Lemma 4.

The immersion $\mathcal{F}$ is determined by the following system of differential equations

$$\begin{array}{cc}\hfill {\mathcal{F}}_{uu}& ={\mathcal{F}}_u,\hfill \\ \hfill {\mathcal{F}}_{uv}& =(\alpha -\delta ){\mathcal{F}}_u+{\mathcal{F}}_v-\mathcal{F},\hfill \\ \hfill {\mathcal{F}}_{vv}& =2\alpha (\delta -\alpha ){\mathcal{F}}_u+(\delta -\alpha ){\mathcal{F}}_v+2\alpha \mathcal{F}.\hfill \end{array}$$

Proof. 

Recall that ${\widehat{\nabla }}_{E_1}{E}_1={\widehat{\nabla }}_{E_1}{E}_2=0$, ${\widehat{\nabla }}_{E_2}{E}_2=\delta E_2$ and ${\widehat{\nabla }}_{E_2}{E}_1=-\delta E_1$.

We find

$${\mathcal{F}}_{uu}=0+d\mathcal{F}\left(E_1\right)-0={\mathcal{F}}_u,$$

$$\begin{array}{cc}\hfill {\mathcal{F}}_{uv}& \hfill =d\mathcal{F}\left({\tilde{\nabla }}_{E_1}(E_2-\alpha E_1)\right)+d\mathcal{F}\left(K(E_1,E_2-\alpha E_1)\right)-h(E_1,E_2-\alpha E_1)\mathcal{F}\\ \hfill & =-E_1\left(\alpha \right)d\mathcal{F}\left(E_1\right)+d\mathcal{F}(\alpha E_1+E_2-\alpha E_1)-\mathcal{F}\hfill \\ \hfill & =-\delta {\mathcal{F}}_u+\alpha {\mathcal{F}}_u+{\mathcal{F}}_v-\mathcal{F},\hfill \end{array}$$

and

$$\begin{array}{c}{\mathcal{F}}_{vv}=d\mathcal{F}\left({\tilde{\nabla }}_{E_2}(E_2-\alpha E_1)\right)-\alpha d\mathcal{F}\left({\tilde{\nabla }}_{E_1}(E_2-\alpha E_1)\right)+d\mathcal{F}(K(E_2,E_2)-2\alpha K(E_1,E_2)+{\alpha }^{2}K(E_1,E_1))+2\alpha \mathcal{F}\hfill \\ =\delta d\mathcal{F}\left(E_2\right)-E_2\left(\alpha \right)d\mathcal{F}\left(E_1\right)+\alpha \delta d\mathcal{F}\left(E_1\right)+\alpha E_1\left(\alpha \right)d\mathcal{F}\left(E_1\right)+d\mathcal{F}(\alpha E_2-2\alpha (\alpha E_1+E_2)+{\alpha }^2{E}_1)+2\alpha \mathcal{F}\hfill \\ =\delta {\mathcal{F}}_v+2\alpha \delta {\mathcal{F}}_u+d\mathcal{F}(-\alpha E_2+{\alpha }^2{E}_1-2{\alpha }^2{E}_1)+2\alpha \mathcal{F}\hfill \\ =\delta {\mathcal{F}}_v+2\alpha \delta {\mathcal{F}}_u-\alpha {\mathcal{F}}_v-2{\alpha }^2{\mathcal{F}}_u+2\alpha \mathcal{F}\hfill \\ =2\alpha (\delta -\alpha ){\mathcal{F}}_u+(\delta -\alpha ){\mathcal{F}}_v+2\alpha \mathcal{F}.\hfill \end{array}$$

□

From the first equation, in all cases, we deduce that there exist vector valued functions $a_1$ and $a_2$ such that $\mathcal{F}(u,v)=a_1\left(v\right)exp\left(u\right)+a_2\left(v\right)$.

4.1. Case c = 0

Using the Corollary 1 and the derivatives of $\alpha$ and $\delta$ in the direction of u and v, we get $\delta =exp\left(u\right)$ and $\alpha =exp\left(u\right)+1$ or $\delta =exp(-u)$ and $\alpha =-exp(-u)+1$.

4.1.1. Case $\delta =exp\left(u\right)$ and $\alpha =exp\left(u\right)+1$

The equations of the Lemma 4 become

$$\begin{array}{cc}\hfill {\mathcal{F}}_{uu}& ={\mathcal{F}}_u,\hfill \\ \hfill {\mathcal{F}}_{uv}& ={\mathcal{F}}_u+{\mathcal{F}}_v-\mathcal{F},\hfill \\ \hfill {\mathcal{F}}_{vv}& =-2(exp\left(u\right)+1){\mathcal{F}}_u-{\mathcal{F}}_v+2(exp\left(u\right)+1)\mathcal{F}.\hfill \end{array}$$

We write $\mathcal{F}(u,v)=a_1\left(v\right)exp\left(u\right)+a_2\left(v\right)$.

The second equation gives $a_1^{\prime }\left(v\right)exp\left(u\right)=a_1\left(v\right)exp\left(u\right)+a_1^{\prime }\left(v\right)exp\left(u\right)+a_2^{\prime }\left(v\right)-a_1\left(v\right)exp\left(u\right)-a_2\left(v\right)$ so

$$a_2^{\prime }\left(v\right)=a_2\left(v\right).$$

Then there exists a constant vector $C_1$ such that $a_2\left(v\right)=C_{1}exp\left(v\right)$.

The third equation gives

$$\begin{array}{cc}\hfill a{″}_{1}exp\left(u\right)″a{″}_2\left(v\right)=& -2(exp\left(u\right)+1)a_1\left(v\right)exp\left(u\right)-a_1^{\prime }\left(v\right)exp\left(u\right)-a_2^{\prime }\left(v\right)\hfill \\ & +2(exp\left(u\right)+1)a_1\left(v\right)exp\left(u\right)+2exp\left(u\right)a_2\left(v\right)+2a_2\left(v\right).\hfill \end{array}$$

So $a{″}_1\left(v\right)exp\left(u\right)=-a_1^{\prime }\left(v\right)exp\left(u\right)+2a_2\left(v\right)exp\left(u\right)$ and finally we get

$$a{″}_1\left(v\right)=-a_1^{\prime }\left(v\right)+2a_2\left(v\right).$$

Then there exist constants vectors C, $C_2$ and $C_3$ such that $a_1=Cexp\left(v\right)+C_{2}exp(-v)+C_3$ and using the previous equation, we get $C=C_1$.

We can use an affine transformation which maps the basis $C_1,C_2,C_3$ to the standard basis. Then

$$\mathcal{F}=(C_{1}exp\left(v\right)+C_{2}exp(-v)+C_3)exp\left(u\right)+C_{1}exp\left(v\right)$$

where $C_1=(1,0,0)$, $C_2=(0,1,0)$ and $C_3=(0,0,1)$.

4.1.2. Case $\delta =exp(-u)$ and $\alpha =-exp(-u)+1$

The equations of the Lemma 4 become

$$\begin{array}{ll}{\mathcal{F}}_{uu}& ={\mathcal{F}}_u,\\ {\mathcal{F}}_{uv}& =(-2exp(-u)+1){\mathcal{F}}_u+{\mathcal{F}}_v-\mathcal{F},\\ {\mathcal{F}}_{vv}& =2(1-exp(-u))(2exp(-u)-1){\mathcal{F}}_u+(2exp(-u)-1){\mathcal{F}}_v+2(1-exp(-u))\mathcal{F}.\end{array}$$

As in the previous case, we write $\mathcal{F}(u,v)=a_1\left(v\right)exp\left(u\right)+a_2\left(v\right)$. The second equation gives

$$a_2^{\prime }\left(v\right)=a_2\left(v\right)+2a_1\left(v\right).$$

Furthermore, after straightforward computations, we find that the third equation reduces to

$$a{″}_1\left(v\right)=-a_1^{\prime }\left(v\right).$$

So there exist constant vectors $C_1$ and $C_2$ such that $a_1\left(v\right)=C_{1}exp(-v)+C_2$.

Moreover, there exist constant vectors $C_3,C_4$ and $C_5$ such that $a_2\left(v\right)=C_{3}exp\left(v\right)+C_{4}exp(-v)+C_5$ and by using $a_2^{\prime }\left(v\right)=a_2\left(v\right)+2a_1\left(v\right)$, we show that $C_4=-C_1$ and $C_5=-2C_2$

Then $a_1\left(v\right)=C_{1}exp(-v)+C_2$ and $a_2\left(v\right)=C_{3}exp\left(v\right)-C_{1}exp(-v)-2C_2$. By using an affine transformation which maps the basis $C_1,C_2,C_3$ to the standard basis, we have

$$\mathcal{F}=(C_{1}exp(-v)+C_2)exp\left(u\right)+C_{3}exp\left(v\right)-C_{1}exp(-v)-2C_2$$

where $C_1=(1,0,0)$, $C_2=(0,1,0)$ and $C_3=(0,0,1)$.

4.2. Case c = 1

We can suppose that $\alpha -1=rcosh\left(u\right)$ and $\delta =rsinh\left(u\right)$. The equations of the Lemma 4 become

$$\begin{array}{cc}\hfill {\mathcal{F}}_{uu}& ={\mathcal{F}}_u,\hfill \\ \hfill {\mathcal{F}}_{uv}& =(rcosh\left(u\right)-rsinh\left(u\right)+1){\mathcal{F}}_u+{\mathcal{F}}_v-\mathcal{F},\hfill \\ \hfill {\mathcal{F}}_{vv}& =2(rcosh\left(u\right)+1)(rsinh\left(u\right)-rcosh\left(u\right)-1){\mathcal{F}}_u+(rsinh\left(u\right)-rcosh\left(u\right)-1){\mathcal{F}}_v+2(rcosh\left(u\right)+1)\mathcal{F}.\hfill \end{array}$$

We write $\mathcal{F}(u,v)=a_1\left(v\right)exp\left(u\right)+a_2\left(v\right)$.

The second equation gives

$$a_2\left(v\right)=a_2^{\prime }\left(v\right)+r{a}_1\left(v\right).$$

Furthermore, after straightforward computations, we find that the third equation reduces to $a{″}_1\left(v\right)exp\left(u\right)=-a_1^{\prime }\left(v\right)exp\left(u\right)+r{a}_2^{\prime }\left(v\right)exp\left(u\right)$. Therefore we get the relation

$$a{″}_1\left(v\right)=-a_1^{\prime }\left(v\right)+r{a}_2^{\prime }\left(v\right).$$

We have $r{a}_1\left(v\right)=a_2\left(v\right)-a_2^{\prime }\left(v\right)$ and we derive two times. Since $r\ne 0$, we get $ra{″}_1\left(v\right)=-r{a}_1^{\prime }\left(v\right)+r^2{a}_2^{\prime }\left(v\right)$. Then computations give

$$a_2^{\left(3\right)}\left(v\right)=(1-r^2)a_2^{\prime }\left(v\right).$$

4.2.1. Case $1-r^2>0$

Then there exist constants vectors $C_1,C_2$ and $C_3$ such that

$$a_2\left(v\right)=C_{1}exp\left(\sqrt{1-r^2}v\right)+C_{2}exp(-\sqrt{1-r^2}v)+C_3$$

and using the relation $a_2\left(v\right)=a_2^{\prime }\left(v\right)+r{a}_1\left(v\right)$, we get

$$a_1\left(v\right)=C_1\left({\displaystyle \frac{1+\sqrt{1-r^2}}{r}}\right)exp\left(\sqrt{1-r^2}v\right)+C_2\left({\displaystyle \frac{1-\sqrt{1-r^2}}{r}}\right)exp(-\sqrt{1-r^2}v)+{\displaystyle \frac{C_3}{r}}.$$

By using an affine transformation which maps the basis $C_1,C_2,C_3$ to the standard basis, we have

$$\begin{array}{ll}\mathcal{F}=& \left(C_1\left({\displaystyle \frac{1+\sqrt{1-r^2}}{r}}\right)exp\left(\sqrt{1-r^2}v\right)+C_2\left({\displaystyle \frac{1-\sqrt{1-r^2}}{r}}\right)exp(-\sqrt{1-r^2}v)+{\displaystyle \frac{C_3}{r}}\right)exp\left(u\right)\\ & +C_{1}exp\left(\sqrt{1-r^2}v\right)+C_{2}exp(-\sqrt{1-r^2}v)+C_3\end{array}$$

where $C_1=(1,0,0)$, $C_2=(0,1,0)$ and $C_3=(0,0,1)$.

4.2.2. Case $1-r^2<0$

Then there exist constants vectors $C_1,C_2$ and $C_3$ such that

$$a_2\left(v\right)=C_{1}cos\left(\sqrt{r^2-1}v\right)+C_{2}sin\left(\sqrt{r^2-1}v\right)+C_3$$

and using the relation $a_2\left(v\right)=a_2^{\prime }\left(v\right)+r{a}_1\left(v\right)$, we get

$$a_1\left(v\right)=\left({\displaystyle \frac{C_1-C_2\sqrt{r^2-1}}{r}}\right)cos\left(\sqrt{r^2-1}v\right)+\left({\displaystyle \frac{C_2+C_1\sqrt{r^2-1}}{r}}\right)sin\left(\sqrt{r^2-1}v\right)+{\displaystyle \frac{C_3}{r}}.$$

By using an affine transformation which maps the basis $C_1,C_2,C_3$ to the standard basis, we have

$$\begin{array}{ll}\mathcal{F}=& \left(\left({\displaystyle \frac{C_1-C_2\sqrt{r^2-1}}{r}}\right)cos\left(\sqrt{r^2-1}v\right)+\left({\displaystyle \frac{C_2+C_1\sqrt{r^2-1}}{r}}\right)sin\left(\sqrt{r^2-1}v\right)+{\displaystyle \frac{C_3}{r}}\right)exp\left(u\right)\hfill \\ & +C_{1}cos\left(\sqrt{r^2-1}v\right)+C_{2}sin\left(\sqrt{r^2-1}v\right)+C_3\hfill \end{array}$$

where $C_1=(1,0,0)$, $C_2=(0,1,0)$ and $C_3=(0,0,1)$.

4.2.3. Case $r=1$

We have $a_2^{\left(3\right)}\left(v\right)=0$. So there exist constant vectors $C_1,C_2$ and $C_3$ such that

$$a_2\left(v\right)=C_1{\displaystyle \frac{v^2}{2}}+C_{2}v+C_3$$

and using the relation $a_2\left(v\right)=a_2^{\prime }\left(v\right)+a_1\left(v\right)$, we get

$$a_1\left(v\right)=C_1{\displaystyle \frac{v^2}{2}}+(C_2-C_1)v+(C_3-C_2).$$

By using an affine transformation which maps the basis $C_1,C_2,C_3$ to the standard basis, we have

$$\mathcal{F}=\left(C_1{\displaystyle \frac{v^2}{2}}+(C_2-C_1)v+(C_3-C_2)\right)exp\left(u\right)+C_1{\displaystyle \frac{v^2}{2}}+C_{2}v+C_3$$

where $C_1=(1,0,0)$, $C_2=(0,1,0)$ and $C_3=(0,0,1)$.

4.3. Case c = −1

We can suppose that $\alpha -1=rsinh\left(u\right)$ and $\delta =rcosh\left(u\right)$. The equations of the Lemma 4 become

$$\begin{array}{cc}\hfill {\mathcal{F}}_{uu}& ={\mathcal{F}}_u,\hfill \\ \hfill {\mathcal{F}}_{uv}& =(rsinh\left(u\right)-rcosh\left(u\right)+1){\mathcal{F}}_u+{\mathcal{F}}_v-\mathcal{F},\hfill \\ \hfill {\mathcal{F}}_{vv}& =2(rsinh\left(u\right)+1)(rcosh\left(u\right)-rsinh\left(u\right)-1){\mathcal{F}}_u+(rcosh\left(u\right)-rsinh\left(u\right)-1){\mathcal{F}}_v+2(rsinh\left(u\right)+1)\mathcal{F}.\hfill \end{array}$$

As before, $\mathcal{F}(u,v)=a_1\left(v\right)exp\left(u\right)+a_2\left(v\right)$.

The second equation gives

$$a_2\left(v\right)=a_2^{\prime }\left(v\right)-r{a}_1\left(v\right).$$

Furthermore, after straightforward computations, we find that the third equation reduces to $a{″}_1\left(v\right)exp\left(u\right)=-a_1^{\prime }\left(v\right)exp\left(u\right)+r{a}_2^{\prime }\left(v\right)exp\left(u\right)$. Therefore we get the relation

$$a{″}_1\left(v\right)=-a_1^{\prime }\left(v\right)+r{a}_2^{\prime }\left(v\right).$$

We have $r{a}_1\left(v\right)=a_2^{\prime }\left(v\right)-a_2\left(v\right)$ and we derive two times. Since $r\ne 0$, we have $ra{″}_1\left(v\right)=-r{a}_1^{\prime }\left(v\right)+r^2{a}_2^{\prime }\left(v\right)$. Then computations give

$$a_2^{\left(3\right)}\left(v\right)=(1+r^2)a_2^{\prime }\left(v\right).$$

Hence there exist constants vectors $C_1,C_2$ and $C_3$ such that

$$a_2\left(v\right)=C_{1}exp\left(\sqrt{1+r^2}v\right)+C_{2}exp(-\sqrt{1+r^2}v)+C_3$$

and using the relation $r{a}_1\left(v\right)=a_2^{\prime }\left(v\right)-a_2\left(v\right)$, we get

$$a_1\left(v\right)=C_1\left({\displaystyle \frac{\sqrt{1+r^2}-1}{r}}\right)exp\left(\sqrt{1+r^2}v\right)-C_2\left({\displaystyle \frac{\sqrt{1+r^2}+1}{r}}\right)exp(-\sqrt{1+r^2}v)-{\displaystyle \frac{C_3}{r}}.$$

By using an affine transformation which maps the basis $C_1,C_2,C_3$ to the standard basis, we have

$$\begin{array}{ll}\mathcal{F}=& \left(C_1\right({\displaystyle \frac{\sqrt{1+r^2}-1}{r}})exp\left(\sqrt{1+r^2}v\right)-C_2({\displaystyle \frac{\sqrt{1+r^2}+1}{r}})exp(-\sqrt{1+r^2}v)-{\displaystyle \frac{C_3}{r}})exp\left(u\right)\hfill \\ & +C_{1}exp\left(\sqrt{1+r^2}v\right)+C_{2}exp(-\sqrt{1+r^2}v)+C_3\hfill \end{array}$$

where $C_1=(1,0,0)$, $C_2=(0,1,0)$ and $C_3=(0,0,1)$.

We have proved Theorem 2.